Solution:
![x+1=0\\x=-1\\\\2x-4=0\\x=2\\\\\frac{5x^{2}+8x+2}{2x}=0\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}](https://tex.z-dn.net/?f=x%2B1%3D0%5C%5Cx%3D-1%5C%5C%5C%5C2x-4%3D0%5C%5Cx%3D2%5C%5C%5C%5C%5Cfrac%7B5x%5E%7B2%7D%2B8x%2B2%7D%7B2x%7D%3D0%5C%5Cx%3D%5Cfrac%7B-4%2B%5Csqrt%5B%5D%7B6%7D%7D%7B5%7D%5C%5Cx%3D%5Cfrac%7B-4-%5Csqrt%5B%5D%7B6%7D%7D%7B5%7D)
Answer:
![x=-1\\x=2\\x=\frac{-4+\sqrt[]{6}}{5}\\x=\frac{-4-\sqrt[]{6}}{5}](https://tex.z-dn.net/?f=x%3D-1%5C%5Cx%3D2%5C%5Cx%3D%5Cfrac%7B-4%2B%5Csqrt%5B%5D%7B6%7D%7D%7B5%7D%5C%5Cx%3D%5Cfrac%7B-4-%5Csqrt%5B%5D%7B6%7D%7D%7B5%7D)
<em>Hope this was helpful.</em>
For the derivative tests method, assume that the sphere is centered at the origin, and consider the
circular projection of the sphere onto the xy-plane. An inscribed rectangular box is uniquely determined
1
by the xy-coordinate of its corner in the first octant, so we can compute the z coordinate of this corner
by
x2+y2+z2=r2 =⇒z= r2−(x2+y2).
Then the volume of a box with this coordinate for the corner is given by
V = (2x)(2y)(2z) = 8xy r2 − (x2 + y2),
and we need only maximize this on the domain x2 + y2 ≤ r2. Notice that the volume is zero on the
boundary of this domain, so we need only consider critical points contained inside the domain in order
to carry this optimization out.
For the method of Lagrange multipliers, we optimize V(x,y,z) = 8xyz subject to the constraint
x2 + y2 + z2 = r2<span>. </span>
If your directrix is a "y=" line, that means that the parabola opens either upwards or downwards (as opposed to the left or the right). Because it is in the character of a parabola to "hug" the focus, our parabola opens upwards. The vertex of a parabola sits exactly halfway between the directrix and the focus. Since our directrix is at y = -2 and the focus is at (1, 6) AND the parabola opens upward, the vertex is going to sit on the main transversal, which is also the "line" the focus sits on. The focus is on the line x = 1, so the vertex will also have that x coordinate. Halfway between the y points of the directrix and the focus, -2 and 6, respectively, is the y value of 2. So the vertex sits at (1, 2). The formula for this type of parabola is 
where h and k are the coordinates of the vertex and p is the DISTANCE that the focus is from the vertex. Our focus is 4 units from the vertex, so p = 4. Filling in our h, k, and p: 
. Simplifying a bit gives us 
. We can begin to isolate the y by dividing both sides by 16 to get 
. Then we can add 2 to both sides to get the final equation 
, choice 4 from above.
